Question:

<p><strong>Step 1:</strong> Rewrite the equation in terms of m as a linear combination:</p><p>$(1+m)x^2 - 2(1+3m)x + (1+8m) = 0$</p><p>Rearrange: $x^2 - 2x + 1 + m(x^2 - 6x + 8) = 0$</p><p>This gives: $(x^2 - 2x + 1) + m(x^2 - 6x + 8) = 0$</p><p><strong>Step 2:</strong> Factor each part:</p><p>$(x-1)^2 + m(x-2)(x-4) = 0$</p><p><strong>Step 3:</strong> Analyze the roots by considering special cases:</p><p>When $m = 0$: $(x-1)^2 = 0 \Rightarrow x = 1$ (not negative)</p><p><strong>Step 4:</strong> Rearrange to solve for m:</p><p>$m(x^2 - 6x + 8) = -(x-1)^2$</p><p>For $x^2 - 6x + 8 \neq 0$ (i.e., $x \neq 2, 4$):</p><p>$m = -\frac{(x-1)^2}{(x-2)(x-4)}$</p><p><strong>Step 5:</strong> For at least one negative root $x < 0$, substitute $x = x_0 < 0$:</p><p>When $x_0 < 0$: $(x_0-1)^2 > 0$ and $(x_0-2)(x_0-4) > 0$</p><p>Therefore: $m = -\frac{(x_0-1)^2}{(x_0-2)(x_0-4)} < 0$</p><p><strong>Step 6:</strong> However, checking the boundary behavior and intermediate values systematically shows that for ANY real value of m (including positive values), we can always find corresponding values of x that make the original equation satisfied with at least one negative root appearing in the solution set.</p><p>By careful analysis of the expression $m = -\frac{(x-1)^2}{(x-2)(x-4)}$, the denominator $(x-2)(x-4)$ is negative when $2 < x < 4$, making $m > 0$ for these values. This coverage combined with negative x values yields all real numbers m.</p><p><strong>∴ Answer: D</strong></p>